Least squares for a line through the origin

1. $S(a)=(2-a)^2+(3-2a)^2+(7-3a)^2=14a^2-58a+62$. Then $S'(a)=28a-58=0$ gives
$$a=\frac{29}{14}\approx2.07.$$
The minimal value is $S=62-\dfrac{29^2}{14}=\dfrac{27}{14}\approx1.93$.

2. For $n$ points, $S(a)=a^2\sum x_i^2-2a\sum x_iy_i+\sum y_i^2$ and
$$S'(a)=0\iff a=\frac{\sum x_iy_i}{\sum x_i^2}$$
(provided not all $x_i$ are $0$). For the data above, $\sum x_iy_i=29$ and $\sum x_i^2=14$.

Remark. Convexity ($S''=2\sum x_i^2>0$) guarantees that the critical point is the global minimum. With a free intercept ($y=ax+b$) one has two parameters and gets the usual regression line, but the idea is the same.